Question
Two radioactive substances $$A$$ and $$B$$ have decay constants $$5\lambda $$ and $$\lambda $$ respectively. At $$t = 0$$ they have the same number of nuclei. The ratio of number of nuclei of $$A$$ to those of $$B$$ will be $${\left( {\frac{1}{e}} \right)^2}$$ after a time interval
A.
$$\frac{1}{{4\lambda }}$$
B.
$$4\lambda $$
C.
$$2\lambda $$
D.
$$\frac{1}{{2\lambda }}$$
Answer :
$$\frac{1}{{2\lambda }}$$
Solution :
Number of nuclei remained after time $$t$$ can be written as
$$N = {N_0}{e^{ - \lambda t}}$$
where, $${N_0}$$ is initial number of nuclei of both the substances.
$$\eqalign{
& {N_1} = {N_0}{e^{ - 5\lambda t}}\,.......\left( {\text{i}} \right) \cr
& {\text{and}}\,\,{N_2} = {N_0}{e^{ - \lambda t}}\,.......\left( {{\text{ii}}} \right) \cr} $$
Dividing Eq. (i) by Eq. (ii), we obtain
$$\frac{{{N_1}}}{{{N_2}}} = {e^{\left( { - 5\lambda + \lambda } \right)t}} = {e^{ - 4\lambda t}} = \frac{1}{{{e^{4\lambda t}}}}$$
But, we have given
$$\frac{{{N_1}}}{{{N_2}}} = {\left( {\frac{1}{e}} \right)^2} = \frac{1}{{{e^2}}}$$
Hence, $$\frac{1}{{{e^2}}} = \frac{1}{{{e^{4\lambda t}}}}$$
Comparing the powers, we get
$$2 = 4\lambda t\,\,{\text{or}}\,\,t = \frac{2}{{4\lambda }} = \frac{1}{{2\lambda }}$$